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How do you differentiate # y =-ln [ 3+ (9+x^2) / x]#?

1 Answer

Dec 24, 2017

#dy/dx=-1/(3+9/x+x)*(1-9/x^2)#

Explanation:

#y=-ln(3+(9+x^2)/x)=-ln(3+9/x+x)#

Use the chain rule #dy/dx=dy/(du)*(du)/dx#

Let #y=-ln(u)# then #dy/(du)=-1/u#
And #u=3+9/x+x# then #(du)/(dx)=-9/x^2+1#

#dy/dx=-1/u*(1-9/x^2)#

Substitute #u=3+9/x+x#

#dy/dx=-1/(3+9/x+x)*(1-9/x^2)#

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